Celestial Mechanics: Newton's Law of Universal Gravitation
CelestialMechanicsNewtonsLawOfUniversalGravitation
2026-09-20 02:21:45
2026-09-20 02:21:45
Topic
added computational resource
Newtonian gravity
universal gravitation
inverse square law
gravitational field
superposition
gravitational constant
point mass
spherical symmetry
gravitational parameter
two-body force
celestial mechanics
astrophysics
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\section*{Celestial Mechanics: Newton's Law of Universal Gravitation}
Celestial mechanics begins with a force law. Before discussing ellipses, orbital elements, Kepler's equation, perturbations, or three-body motion, we first need a precise mathematical statement of how two masses attract one another.
Newton's law of universal gravitation says that two point masses $m_1$ and $m_2$, separated by distance $r$, attract each other with force magnitude
\begin{equation}
\boxed{
F=G\frac{m_1m_2}{r^2}.}
\end{equation}
This compact expression contains several distinct pieces of physics:
\begin{itemize}
\item the interaction is proportional to each mass;
\item it becomes weaker with the square of separation;
\item the force acts along the line joining the masses;
\item the force is attractive;
\item the two bodies exert equal and opposite forces on one another.
\end{itemize}
The purpose of CM01 is to unpack those statements carefully and express them in the vector language needed for later celestial mechanics. The lesson also introduces the gravitational field, superposition, continuous mass distributions, the gravitational parameter, and the first astronomical-scale calculations. The potential formulation is previewed here and developed systematically in CM02 \cite{Newton1999,Goldstein2001,Danby1988,CarrollOstlie2007}.
\section{From a scalar force magnitude to a vector law}
The scalar expression
\begin{equation}
F=G\frac{m_1m_2}{r^2}
\end{equation}
tells us the size of the force but not its direction. Orbital motion requires a vector equation.
Let the inertial positions of the masses be
\begin{equation}
\mathbf r_1
\qquad\text{and}\qquad
\mathbf r_2.
\end{equation}
Define the relative displacement from body 1 to body 2 by
\begin{equation}
\boxed{
\mathbf r=\mathbf r_2-\mathbf r_1.}
\end{equation}
Its magnitude is
\begin{equation}
r=|\mathbf r|,
\end{equation}
and the corresponding unit vector is
\begin{equation}
\widehat{\mathbf r}=\frac{\mathbf r}{r}.
\end{equation}
The force on body 2 due to body 1 must point from body 2 back toward body 1, opposite to $\widehat{\mathbf r}$. Therefore
\begin{equation}
\boxed{
\mathbf F_{2\leftarrow1}
=-G\frac{m_1m_2}{r^2}\widehat{\mathbf r}.}
\end{equation}
Using
\begin{equation}
\widehat{\mathbf r}=\frac{\mathbf r}{r},
\end{equation}
we obtain the form used throughout celestial mechanics:
\begin{equation}
\boxed{
\mathbf F_{2\leftarrow1}
=-G\frac{m_1m_2}{r^3}\mathbf r.}
\end{equation}
The factor $r^3$ in the denominator sometimes looks surprising because the familiar scalar law contains $r^2$. There is no contradiction. The vector $\mathbf r$ contributes one factor of length:
\begin{equation}
\left|\frac{\mathbf r}{r^3}\right|
=\frac{r}{r^3}
=\frac{1}{r^2}.
\end{equation}
\section{Attraction and the minus sign}
The minus sign is geometric, not decorative.
By definition,
\begin{equation}
\mathbf r=\mathbf r_2-\mathbf r_1
\end{equation}
points from body 1 toward body 2. Gravity on body 2 points the opposite way, toward body 1. Therefore its direction is
\begin{equation}
-\widehat{\mathbf r}.
\end{equation}
If instead one defines the relative vector in the opposite direction, the algebraic sign in the force law changes accordingly. What is physically invariant is that the gravitational force points toward the attracting mass.
This is why defining the relative position vector explicitly is essential before writing a vector force law.
\section{Newton's third law for the pair}
The force on body 1 due to body 2 points from body 1 toward body 2:
\begin{equation}
\boxed{
\mathbf F_{1\leftarrow2}
=+G\frac{m_1m_2}{r^3}\mathbf r.}
\end{equation}
Thus
\begin{equation}
\boxed{
\mathbf F_{1\leftarrow2}
=-\mathbf F_{2\leftarrow1}.}
\end{equation}
This is Newton's third law for the isolated gravitational pair.
\begin{center}
\includegraphics{CM01_fig01_two_body_force_vectors.png}
\vspace{0.45em}
\textbf{Figure.}
With $\mathbf r=\mathbf r_2-\mathbf r_1$, gravity pulls each body toward the other. The two internal forces have equal magnitude and opposite direction.
\end{center}
The third-law structure becomes important in CM08, where the exact two-body problem is separated into center-of-mass motion and relative motion.
\section{The gravitational constant $G$}
The constant $G$ sets the strength of Newtonian gravitation. In SI units,
\begin{equation}
G\approx6.67430\times10^{-11}
\frac{\text{m}^3}{\text{kg}\,\text{s}^2}.
\end{equation}
Unlike the speed of light in SI, $G$ is not fixed by definition; it is determined experimentally.
Its dimensions can be recovered from the force law. Since
\begin{equation}
[F]=\text{kg}\,\text{m}/\text{s}^2,
\end{equation}
we require
\begin{align}
[G]
&=\frac{[F][r]^2}{[m]^2}\\
&=\frac{(\text{kg}\,\text{m}/\text{s}^2)(\text{m}^2)}{\text{kg}^2}\\
&=\boxed{\frac{\text{m}^3}{\text{kg}\,\text{s}^2}}.
\end{align}
This dimensional check is useful because celestial-mechanics formulas often contain the combination $GM$, whose units are
\begin{equation}
[GM]=\frac{\text{m}^3}{\text{s}^2}.
\end{equation}
That combination will soon become so common that it receives its own symbol.
\section{The gravitational parameter}
For a dominant central mass $M$, define the gravitational parameter
\begin{equation}
\boxed{\mu=GM.}
\end{equation}
Then the gravitational acceleration due to a point mass can be written compactly as
\begin{equation}
\mathbf g
=-\frac{\mu}{r^3}\mathbf r.
\end{equation}
In the exact relative two-body problem, the corresponding parameter becomes
\begin{equation}
\mu=G(m_1+m_2),
\end{equation}
which will be derived later rather than assumed.
Using $\mu$ is often preferable in practical orbital work because planetary and stellar gravitational parameters can be determined very accurately from dynamical observations.
\section{Why an inverse-square dependence?}
Newtonian gravity has the radial dependence
\begin{equation}
F\propto\frac{1}{r^2}.
\end{equation}
There is an important geometrical reason that inverse-square laws appear naturally in three-dimensional space.
The area of a sphere of radius $r$ is
\begin{equation}
A(r)=4\pi r^2.
\end{equation}
If some conserved radial flux is distributed uniformly across spherical surfaces, then the flux per unit area must scale as
\begin{equation}
\frac{1}{4\pi r^2}.
\end{equation}
At twice the radius,
\begin{equation}
A(2r)=4\pi(2r)^2=4A(r),
\end{equation}
so the same total flux is spread over four times the area.
\begin{center}
\includegraphics{CM01_fig02_inverse_square_geometry.png}
\vspace{0.45em}
\textbf{Figure.}
The area of a sphere grows as $r^2$, so any conserved spherically symmetric radial flux has a density proportional to $1/r^2$. This explains the geometry associated with an inverse-square field, but does not by itself derive Newton's gravitational law.
\end{center}
That last distinction matters. Geometry explains why a conserved radial flux has inverse-square behavior. It does not prove that gravity must be such a flux or determine the coupling strength $G$. Newtonian gravitation is a physical law supported by observation; the spherical-area argument helps us understand its spatial structure.
\section{A simple scaling consequence}
If the separation changes from $r$ to $2r$, the force changes from
\begin{equation}
F(r)=G\frac{m_1m_2}{r^2}
\end{equation}
to
\begin{align}
F(2r)
&=G\frac{m_1m_2}{(2r)^2}\\
&=\frac{1}{4}F(r).
\end{align}
Thus
\begin{equation}
\boxed{r\rightarrow2r\quad\Longrightarrow\quad F\rightarrow\frac{F}{4}.}
\end{equation}
If the separation triples,
\begin{equation}
F(3r)=\frac{1}{9}F(r).
\end{equation}
This rapid weakening with distance is central to nearly every approximation used later in celestial mechanics.
\section{From force to gravitational field}
It is often useful to separate the property of the source from the mass of the object used to probe it.
Suppose a source mass $M$ is located at the origin and a test mass $m$ is at position $\mathbf r$. The gravitational force on the test mass is
\begin{equation}
\mathbf F
=-G\frac{Mm}{r^3}\mathbf r.
\end{equation}
Define the gravitational field as force per unit test mass:
\begin{equation}
\boxed{
\mathbf g=\frac{\mathbf F}{m}.}
\end{equation}
Therefore
\begin{equation}
\boxed{
\mathbf g(\mathbf r)
=-G\frac{M}{r^3}\mathbf r.}
\end{equation}
Equivalently,
\begin{equation}
\boxed{
\mathbf g(\mathbf r)
=-\frac{GM}{r^2}\widehat{\mathbf r}.}
\end{equation}
Its units are
\begin{equation}
[\mathbf g]=\text{m/s}^2.
\end{equation}
Thus the gravitational field is also an acceleration field.
\section{Why the test mass cancels}
Newton's second law gives
\begin{equation}
\mathbf F=m\mathbf a.
\end{equation}
But gravitational force is
\begin{equation}
\mathbf F=m\mathbf g.
\end{equation}
Therefore
\begin{equation}
m\mathbf a=m\mathbf g.
\end{equation}
For nonzero $m$,
\begin{equation}
\boxed{\mathbf a=\mathbf g.}
\end{equation}
In Newtonian gravity, the acceleration produced by a given external gravitational field is independent of the test body's mass.
\begin{center}
\includegraphics{CM01_fig04_field_vs_force.png}
\vspace{0.45em}
\textbf{Figure.}
The source mass creates the gravitational field. Multiplying the field by a test mass gives the force, and Newton's second law then cancels that test mass, leaving $\mathbf a=\mathbf g$.
\end{center}
This cancellation is the Newtonian manifestation of the equivalence between inertial and gravitational mass. General relativity later elevates this observation into a much deeper geometrical principle, but Newtonian celestial mechanics only requires the result $\mathbf a=\mathbf g$.
\section{Example 1: gravitational acceleration at Earth's surface}
Treat Earth as spherically symmetric and use
\begin{equation}
M_E\approx5.9722\times10^{24}\,\text{kg},
\end{equation}
\begin{equation}
R_E\approx6.371\times10^6\,\text{m}.
\end{equation}
The Newtonian surface gravitational acceleration is
\begin{equation}
g_E=G\frac{M_E}{R_E^2}.
\end{equation}
Substituting,
\begin{align}
g_E
&=(6.67430\times10^{-11})
\frac{5.9722\times10^{24}}{(6.371\times10^6)^2}\\
&\approx9.82\,\text{m/s}^2.
\end{align}
Thus
\begin{equation}
\boxed{g_E\approx9.82\,\text{m/s}^2.}
\end{equation}
The familiar near-surface value of $g$ is therefore not a separate law. It is the local value of Earth's gravitational field.
Real measured surface gravity varies because Earth rotates, is oblate rather than perfectly spherical, has altitude variations, and possesses nonuniform mass distribution. Those refinements belong to gravitational-field modeling rather than the ideal point-mass law.
\section{Example 2: the Sun's gravitational acceleration at Earth}
Let
\begin{equation}
M_\odot\approx1.98847\times10^{30}\,\text{kg}
\end{equation}
and take one astronomical unit as
\begin{equation}
1\,\text{AU}
\approx1.495978707\times10^{11}\,\text{m}.
\end{equation}
The Sun's gravitational field magnitude at Earth's orbital distance is
\begin{align}
g_\odot(1\,\text{AU})
&=G\frac{M_\odot}{r^2}\\
&\approx5.93\times10^{-3}\,\text{m/s}^2.
\end{align}
Therefore
\begin{equation}
\boxed{
g_\odot(1\,\text{AU})\approx5.93\times10^{-3}\,\text{m/s}^2.}
\end{equation}
This is much smaller than Earth's surface gravity, but it acts continuously over astronomical distances and is sufficient to curve Earth's motion around the Sun.
Multiplying by Earth's mass gives the approximate Sun--Earth force magnitude:
\begin{equation}
F_{\odot E}
=M_Eg_\odot
\approx3.54\times10^{22}\,\text{N}.
\end{equation}
The enormous force is accompanied by an enormous planetary mass, so the resulting acceleration remains only a few millimeters per second squared.
\section{Superposition}
Newtonian gravity is linear in the source masses. If several masses are present, the net gravitational force is the vector sum of the individual forces.
For a test mass $m$ at position $\mathbf r$ and source masses $M_i$ at positions $\mathbf r_i$,
\begin{equation}
\boxed{
\mathbf F(\mathbf r)
=\sum_i\mathbf F_i(\mathbf r).}
\end{equation}
The contribution from source $i$ is
\begin{equation}
\mathbf F_i(\mathbf r)
=-GmM_i
\frac{\mathbf r-\mathbf r_i}
{|\mathbf r-\mathbf r_i|^3}.
\end{equation}
Therefore
\begin{equation}
\boxed{
\mathbf F(\mathbf r)
=-Gm\sum_iM_i
\frac{\mathbf r-\mathbf r_i}
{|\mathbf r-\mathbf r_i|^3}.}
\end{equation}
Dividing by the test mass gives the gravitational field:
\begin{equation}
\boxed{
\mathbf g(\mathbf r)
=-G\sum_iM_i
\frac{\mathbf r-\mathbf r_i}
{|\mathbf r-\mathbf r_i|^3}.}
\end{equation}
\begin{center}
\includegraphics{CM01_fig03_superposition.png}
\vspace{0.45em}
\textbf{Figure.}
For several gravitating sources, each mass contributes a vector field at the observation point. The net Newtonian gravitational field is their vector sum.
\end{center}
This superposition principle is the starting point for the Newtonian $N$-body equations developed much later in the series.
\section{A two-source example on one line}
Consider two equal source masses $M$ fixed at
\begin{equation}
x=-a
\qquad\text{and}\qquad
x=+a.
\end{equation}
At the midpoint $x=0$, the left mass produces a field of magnitude
\begin{equation}
\frac{GM}{a^2}
\end{equation}
pointing left, while the right mass produces the same magnitude pointing right. Therefore
\begin{equation}
\boxed{\mathbf g(0)=\mathbf 0.}
\end{equation}
Zero net field does not mean gravity is absent. It means the individual contributions cancel at that point.
The same distinction becomes important in multi-body celestial mechanics, where equilibrium points arise from balances among several gravitational and inertial terms.
\section{From discrete masses to a continuous mass distribution}
If matter is described by a mass density
\begin{equation}
\rho(\mathbf r'),
\end{equation}
then a small source volume $dV'$ contains mass
\begin{equation}
dm=\rho(\mathbf r')\,dV'.
\end{equation}
The field contribution at observation point $\mathbf r$ is
\begin{equation}
d\mathbf g
=-G\rho(\mathbf r')
\frac{\mathbf r-\mathbf r'}{|\mathbf r-\mathbf r'|^3}
\,dV'.
\end{equation}
Integrating over the source gives
\begin{equation}
\boxed{
\mathbf g(\mathbf r)
=-G\int
\rho(\mathbf r')
\frac{\mathbf r-\mathbf r'}{|\mathbf r-\mathbf r'|^3}
\,dV'.}
\end{equation}
This equation extends the point-mass law to planets, stars, gas clouds, stellar systems, and galaxies within Newtonian gravity.
Its direct evaluation can be difficult, but symmetry can simplify it dramatically.
\section{Preview: spherical symmetry and the shell theorem}
For a spherically symmetric body, Newton's shell theorem gives two powerful results:
\begin{enumerate}
\item outside a spherical shell, the gravitational field is exactly the same as if the shell's mass were concentrated at its center;
\item inside an ideal thin spherical shell, the net gravitational field is zero.
\end{enumerate}
Consequently, outside any spherically symmetric body of total mass $M$,
\begin{equation}
\boxed{
\mathbf g
=-G\frac{M}{r^3}\mathbf r,}
\end{equation}
provided $r$ lies outside the mass distribution.
This is why planets and stars can often be treated as point masses when studying external orbital motion, even though they are physically extended objects.
The shell theorem and extended spherical distributions will be developed more fully in CM02E1 when gravitational potential provides a particularly efficient route to the calculation.
\section{What the point-mass approximation really means}
A body does not have to be physically tiny to behave gravitationally like a point mass.
For a spherically symmetric object, the external field is exactly equivalent to that of a point mass at the center. For a nearly spherical body, the point-mass model is often the dominant term in a more detailed gravitational-field expansion.
Thus the approximation
\begin{equation}
\mathbf g\approx-G\frac{M}{r^3}\mathbf r
\end{equation}
can be excellent even when the body's physical radius is large, provided the observer is outside the body and departures from spherical symmetry are unimportant for the required accuracy.
Later perturbation theory will explicitly add nonspherical gravitational effects rather than hiding them inside the ideal law.
\section{Example 3: Earth acting on the Moon}
Take the mean Earth--Moon separation as approximately
\begin{equation}
r_{EM}=3.844\times10^8\,\text{m}.
\end{equation}
The acceleration produced by Earth at that distance is
\begin{align}
g_E(r_{EM})
&=G\frac{M_E}{r_{EM}^2}\\
&\approx2.70\times10^{-3}\,\text{m/s}^2.
\end{align}
Thus
\begin{equation}
\boxed{
g_E(r_{EM})\approx2.70\times10^{-3}\,\text{m/s}^2.}
\end{equation}
Using a lunar mass of approximately
\begin{equation}
M_M\approx7.342\times10^{22}\,\text{kg},
\end{equation}
the mutual force magnitude is approximately
\begin{equation}
\boxed{
F_{EM}\approx1.98\times10^{20}\,\text{N}.}
\end{equation}
Earth experiences exactly the same force magnitude in the opposite direction. Because Earth's mass is much larger, Earth's acceleration toward the Moon is much smaller than the Moon's acceleration toward Earth.
This observation is the first hint that saying ``the Moon orbits Earth'' is an approximation. In the exact two-body problem, both bodies move around their common center of mass.
\section{Gravitational force is central}
For a source at the origin,
\begin{equation}
\mathbf F
=-G\frac{Mm}{r^3}\mathbf r.
\end{equation}
The force is parallel or antiparallel to $\mathbf r$. Therefore it is a central force.
The torque about the force center is
\begin{align}
\boldsymbol\tau
&=\mathbf r\times\mathbf F\\
&=\mathbf r\times
\left(-G\frac{Mm}{r^3}\mathbf r\right)\\
&=\mathbf 0.
\end{align}
Thus
\begin{equation}
\frac{d\mathbf L}{dt}=\mathbf 0.
\end{equation}
We will derive the consequences carefully later, but already we can preview one of the deepest structural facts of celestial mechanics:
\begin{equation}
\boxed{
\text{central gravity}
\Longrightarrow
\text{zero torque about the center}
\Longrightarrow
\text{angular-momentum conservation}.}
\end{equation}
From that conservation law will come planar motion and Kepler's equal-area law.
\section{Preview: gravity is conservative}
The Newtonian point-mass force also has the special property that it can be derived from a scalar potential energy.
For two point masses,
\begin{equation}
\boxed{
U(r)=-\frac{Gm_1m_2}{r},}
\end{equation}
with the conventional choice
\begin{equation}
U(\infty)=0.
\end{equation}
The radial force is recovered from
\begin{equation}
F_r=-\frac{dU}{dr}.
\end{equation}
Indeed,
\begin{align}
-\frac{d}{dr}
\left(-\frac{Gm_1m_2}{r}\right)
&=-\frac{Gm_1m_2}{r^2}.
\end{align}
CM02 will develop gravitational potential and potential energy in detail, including the relation
\begin{equation}
\boxed{\mathbf g=-\nabla\Phi.}
\end{equation}
For now, the important preview is that Newtonian gravity conserves mechanical energy for an isolated system.
\section{Why the potential is negative}
With the reference choice
\begin{equation}
U(\infty)=0,
\end{equation}
two attracting masses at finite separation have
\begin{equation}
U(r)<0.
\end{equation}
The negative sign encodes the fact that positive work must be supplied to separate a bound pair from finite separation to infinity.
This sign will later become central to orbital classification:
\begin{equation}
\text{negative total orbital energy}
\Longrightarrow
\text{bound Kepler orbit}.
\end{equation}
That result belongs later in the sequence, after the kinetic and potential terms have been combined correctly.
\section{Astrophysical scale invariance of the same law}
One of the remarkable features of Newtonian gravity is that the same mathematical law describes systems across an enormous range of scale.
The same expression
\begin{equation}
F=G\frac{m_1m_2}{r^2}
\end{equation}
is used to model, at the appropriate approximation level:
\begin{itemize}
\item a falling laboratory mass near Earth;
\item the Earth--Moon interaction;
\item planets orbiting the Sun;
\item binary stars;
\item stars orbiting a compact central mass;
\item the pairwise forces in an $N$-body star-cluster calculation.
\end{itemize}
The objects and distances change by many orders of magnitude, but the Newtonian force law keeps the same structure until relativistic or extended-mass effects become important.
This universality is one reason celestial mechanics provides such a powerful bridge between classical mechanics and astrophysics.
\section{When Newtonian point-mass gravity is not enough}
The simple law is foundational, but it is not universally sufficient.
Corrections or more detailed models may be required when:
\begin{itemize}
\item the source is significantly nonspherical and its multipole gravity matters;
\item several bodies produce comparable gravitational accelerations;
\item tidal effects depend on the spatial gradient of gravity across an extended body;
\item mass distributions are continuous rather than point-like;
\item velocities or gravitational fields become relativistically strong;
\item nongravitational forces such as radiation pressure, drag, or thrust are important.
\end{itemize}
Celestial mechanics is therefore not about using the most complicated force model possible. It is about selecting the simplest model that contains the physics relevant to the question.
\section{A compact vector summary}
For two point masses with
\begin{equation}
\mathbf r=\mathbf r_2-\mathbf r_1,
\end{equation}
the gravitational force on body 2 due to body 1 is
\begin{equation}
\boxed{
\mathbf F_{2\leftarrow1}
=-Gm_1m_2\frac{\mathbf r}{r^3}.}
\end{equation}
The force on body 1 due to body 2 is
\begin{equation}
\boxed{
\mathbf F_{1\leftarrow2}
=+Gm_1m_2\frac{\mathbf r}{r^3}.}
\end{equation}
For a source mass $M$ at the origin, the gravitational field is
\begin{equation}
\boxed{
\mathbf g(\mathbf r)
=-GM\frac{\mathbf r}{r^3}.}
\end{equation}
The force on a test mass $m$ is
\begin{equation}
\boxed{\mathbf F=m\mathbf g.}
\end{equation}
For many point sources,
\begin{equation}
\boxed{
\mathbf g(\mathbf r)
=-G\sum_iM_i
\frac{\mathbf r-\mathbf r_i}
{|\mathbf r-\mathbf r_i|^3}.}
\end{equation}
For a continuous density distribution,
\begin{equation}
\boxed{
\mathbf g(\mathbf r)
=-G\int
\rho(\mathbf r')
\frac{\mathbf r-\mathbf r'}{|\mathbf r-\mathbf r'|^3}
\,dV'.}
\end{equation}
These equations are the gravitational foundation for the rest of the celestial-mechanics series.
\section{Common mistakes}
\begin{itemize}
\item Writing $1/r^3$ in the vector law and concluding that gravity is an inverse-cube force. The numerator contains the vector $\mathbf r$, so the magnitude still scales as $1/r^2$.
\item Using a minus sign without defining the relative-position vector. The sign only has meaning relative to a coordinate convention.
\item Treating $G$ and $g$ as the same quantity. $G$ is the universal gravitational constant; $\mathbf g$ is a gravitational field or acceleration.
\item Treating one body as fixed without checking the mass ratio. The exact two-body problem allows both masses to move.
\item Adding gravitational magnitudes instead of vectors when more than one source is present.
\item Assuming the inverse-square geometry alone proves Newton's gravitational law. It explains the spherical spreading structure but not the existence or strength of the interaction.
\item Applying the point-mass law inside an arbitrary extended body. The external point-mass equivalence requires spherical symmetry; interior fields require integrating the mass distribution or using symmetry results.
\item Forgetting that $\mathbf a=\mathbf g$ follows after dividing by the test mass. The gravitational force itself is still proportional to the test mass.
\end{itemize}
\section{What CM01 adds to the series}
CM00 gave the roadmap. CM01 supplies the first dynamical law on which that roadmap rests.
The main chain is
\begin{equation}
\boxed{
\text{source masses}
\longrightarrow
\text{Newtonian gravitational force}
\longrightarrow
\text{gravitational field}
\longrightarrow
\text{equations of motion}.}
\end{equation}
The key result for a point source is
\begin{equation}
\boxed{
\mathbf g(\mathbf r)
=-\frac{GM}{r^3}\mathbf r.}
\end{equation}
The next companion entry, CM01E1, should turn this law into worked calculations involving force ratios, surface gravity, two-body force pairs, vector superposition, and astronomical scaling.
The next main theory article, CM02, will develop gravitational potential and potential energy from the same inverse-square force and establish the conservative-field structure
\begin{equation}
\boxed{\mathbf g=-\nabla\Phi.}
\end{equation}
\begin{thebibliography}{9}
\bibitem{CarrollOstlie2007}
Bradley W. Carroll and Dale A. Ostlie,
\emph{An Introduction to Modern Astrophysics},
2nd ed., Pearson/Addison-Wesley, 2007.
\bibitem{Goldstein2001}
Herbert Goldstein, Charles Poole, and John Safko,
\emph{Classical Mechanics},
3rd ed., Addison-Wesley, 2001.
\bibitem{Danby1988}
J. M. A. Danby,
\emph{Fundamentals of Celestial Mechanics},
2nd ed., Willmann-Bell, 1988.
\bibitem{BateMuellerWhite1971}
Roger R. Bate, Donald D. Mueller, and Jerry E. White,
\emph{Fundamentals of Astrodynamics},
Dover Publications, 1971.
\bibitem{MurrayDermott1999}
Carl D. Murray and Stanley F. Dermott,
\emph{Solar System Dynamics},
Cambridge University Press, 1999.
\bibitem{Newton1999}
Isaac Newton,
\emph{The Principia: Mathematical Principles of Natural Philosophy},
trans. I. Bernard Cohen and Anne Whitman,
University of California Press, 1999.
\end{thebibliography}